prob5sol - ISMT 111 Business Statistics Simplified Solution...

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- 1 - ISMT 111 Business Statistics Simplified Solution for Problem Sheet 5 1. H 0 : m 1 - m 2 = 0 H 1 : m 1 - m 2 0 t = 4.19 and t α /2, n1 + n2 2 = t 0.005, 9 = 3.25. Since t > t 0.005, 9 , reject H 0 at 0.05 level of significance. 2. H 0 : m = $33.99 H 1 : m > $33.99 z =2; p -value = P( z > 2) = 0.0228 The range is 1 a 0.0228 3. (a) H 0 : m d = 0 H 1 : m d > 0 t = 95 . 1 8 / 44 . 5 75 . 3 / = = n s d d and t α , n 1 = t 0.025, 7 = 2.365. We cannot reject H 0 . (b) As t 0.025, 7 = 2.365 & t 0.05, 7 = 1.895 0.025 < p-value < 0.05 Hence, the lower bound on the p-value is 0.025 4. (a) H 0 : p = 0.5 H 1 : p > 0.5 z = 1 and z α = z 0.05 = 1.645 Since z < z 0.05 , cannot reject H 0 at 0.05 level of significance. (b) Let m 1 be the population average monthly salary among those with one job offer. Let m 2 be the population average monthly salary among those with three jobs offer. H 0 : m 1 - m 2 = 0 H 1 : m 1 - m 2 0 z = -2.1 p -value = 2 P( z > 2.1) = 2(0.0179) = 0.0358. Since p -value < a = 0.05, reject H 0 at 0.05 level of significance. 5. (a) Let m 1 be the mean salary per hours of male factory workers. Let m 2 be the mean salary per hours of female factory workers. H 0 : m 1 - m 2 = 1.5 H 1 : m 1 - m 2 > 1.5 t = 1.8836 and t α , n1 + n2 2 = t 0.05, 21 = 1.721. Reject H 0 . (b) Let
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prob5sol - ISMT 111 Business Statistics Simplified Solution...

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